Variation of Parameters — Question 10

Question 10 – Solution

Strategy. Treat the residual as an error forcing and bound its full response, not just sampled values.

Step 1: Compute the forcing error. We have $q'''=1+x$ and $q'=x^2/2+x^3/6$, so $$r=1+x+x^2/2+x^3/6-e^x=-R_4(x),$$ where $R_4=e^x-(1+x+x^2/2+x^3/6)$. The values $q(0),q'(0),q''(0)$ are zero. Thus $E=y-q$ has a zero initial vector and satisfies $E'''+E'=R_4$.

Step 2: Derive the positive error integral. For the basis $1,\cos x,\sin x$, the variation derivatives for this error equation are $(R_4,-R_4\cos x,-R_4\sin x)$. Integrating from zero gives $$\boxed{E(x)=\int_0^x[1-\cos(x-t)]R_4(t)\,dt.}$$ For $0<t<x\le 1$, both factors are strictly positive. Hence $E(x)>0$ on $(0,1]$: the approximation underestimates the true solution.

Step 3: Certify a uniform error bound. Taylor’s theorem gives $0\le R_4(t)\le \mathrm et^4/24$ on $[0,1]$. Therefore $$0\le E(x)\le\frac{\mathrm e}{48}\int_0^x(x-t)^2t^4\,dt =\boxed{\frac{\mathrm ex^7}{5040}},$$ where the polynomial integral equals $x^7/105$. The maximum certified error is $\mathrm e/5040\approx 0.00053934<10^{-3}$, so the tolerance holds throughout $[0,1]$.

Step 4: Compare with an exact independent check. For $Y=(e^x+\cos x-\sin x)/2-1$, direct differentiation gives $Y'''+Y'=e^x$ and its initial vector is zero. Uniqueness therefore implies $Y=y$. At $1$, $q(1)=5/24$, so the actual error is $$\boxed{E(1)=\frac{\mathrm e+\cos 1-\sin 1}{2}-\frac{29}{24}.}$$ Numerically, $E(1)\approx 0.00022324$, positive and smaller than the certified bound. Exact evaluation or sample agreement is useful corroboration, but finitely many values cannot exclude larger errors between them. The integral inequality proves the bound for every point of the interval.